Let $(\textbf{a}_{n})_{n = 1}^{\infty}$ be a sequence in $\mathbb{R}^k$. Show $\{\textbf{a}_{n} : n \geq 1 \} \cup \{\textbf{a}\}$ is closed. Let $(\textbf{a}_{n})_{n = 1}^{\infty}$ be a sequence in $\mathbb{R}^k$ and $$\lim_{n \rightarrow \infty}\textbf{a}_{n} = \textbf{a}.$$ Show $B = \{\textbf{a}_{n} : n \geq 1 \} \cup \{\textbf{a}\}$ is closed.
Edit: I would specifically like to use the limit point definition of closed.  i.e:


A set $A$ is considered closed if it contains all of its limit points


EDIT: I think I've gotten a way to prove it.
Attempt 2:
We will show $B^{c}$ is open. This means I have to construct an open ball $B_{\delta}(x)$ s.t $B_{\delta}(x) \subset B^{c}$.
Given that the sequence $a_{n}$ converges then  for all $\epsilon > 0$ there exists a $N_{\epsilon} > 0$ such that for all $n > N_{\epsilon}$, $\|a_{n} - a \| < \epsilon$.
Let $x \in B^{c}$, $\epsilon > 0$, and $\epsilon = \frac{|x-a|}{2}$.
As $a_{n} \rightarrow a$ then $\|a_{n} - a\| < \frac{|x-a|}{2}$, let $\omega = min \{\frac{|x-a|}{2}, \frac{|x-a_{1}|}{2}, \dots, \frac{|x-a_{n}|}{2}\}$.
This then means $\|x-a_{n}\| > \omega \\ \Rightarrow \ B_{\omega}(x) \subset B^{c}$
Therefore $B^{c}$ is open and this means $B$ is closed.
Comment:
I'm not sure if the ball I constructed makes sense. I see it visually what I would like to accomplish but I may not be articulating it correctly.
 A: Since the sequence converges, the set of the limit points is a subset of $\{\textbf{a}\}$. A convergent sequence has either no limit point (e.g. constant sequence), or only one (otherwise this would be a contradiction to the fact that the sequence converges).
Because $\{\textbf{a}\}\subset B$, $B$ is closed.
A: Suppose there is a different limit point $\mathbf b$ of $\mathbf B$. Set $\varepsilon=\frac12|\mathbf b-\mathbf a|$. Then by convergence of the sequence there are only finitely many points outside $B_ε(\mathbf a)$. 
However at the same time, as $\mathbf b$ is a limit point, there need to be infinitely many points of $\mathbf B$ inside $B_ε(\mathbf b)$, which is contained in that "outside". 
This is a contradiction.
A: Another interesting proof is by showing that $B$ is compact. Suppose that $(V_\alpha)$ is an open cover of $B$. Then $a\in B\Rightarrow a\in V_{\alpha_0}$ for some $\alpha_0$. As $a_n\to a$ and $V_{\alpha_0}$ is open, it follows that exists $n_0$ such that for all $n\geq n_0$ we have $a_n\in V_{\alpha_0}$. For each $n<n_0$ choose $V_{\alpha_n}$ containing $a_n$. Then $(V_{\alpha_n})_{0\leq n<n_0}$ covers $B$, and therefore $B$ is compact, and in particular closed.
A: Let $x\in\Bbb R^n\setminus B$ and let $e=d(x,\mathbf{a})>0$.
Choose $\eta<\frac12e$. Since $\mathbf{a}_n\in S(\mathbf{a},\eta)$ for all $n$ sufficiently large, we must have that 
$$
B\cap S(x,\eta)=\{b_1,...,b_r\}\quad\text{is a finite set}
$$
because $S(\mathbf{a},\eta)\cap S(x,\eta)=\emptyset$.
Since $x\neq\mathbf{a}_n$ we have
$$
\min d(x,b_i)=\delta>0.
$$
Thus
$$
S(x,\delta/2)\cap B=\emptyset.
$$
This shows that $\Bbb R^n\setminus B$ is open.
A: A subset of a metric space is closed if it contains all its limit points. Here the given set contains its limit point $a$ and hence it is closed.
